Question

For each initial value problem, does Picards's theorem apply? If so, determine if it guarantees that a solutio exists and is unique.

Theorem (Picard). Consider the initial value problem dy = f(t,y), dt (IVP) y(to) = Yo- (a) Existence: If f(t,y) is continuous in an open rectangle R = {(t,y) |a<t < b, c < y < d} and (to, Yo) belongs in R, then there exist h > 0 and a solution y = y(t) of (IVP) defined in the interval to - h <t< to + h. (6) Uniqueness: If, in addition, (t,y) is continuous in R, then the solution given in part (a) is unique. Problem: For each initial value problem, does Picard's theorem apply? If so, determine if it guarantees that a solution exists and is unique. (a) (b) ( dy { dt = 2vy – 1, \ y(0) = 2. de = VI - Y, y(1) = 1.

Answer 1

(a) dy/dt = f(t, y) = 2 squareroot y - 1 y(0) = 2 f is continuous on region y greterthanorequalto 1 where y_0 = 2 greterthanorequalto 1 so there exist solution f_y = 2 1/2 squareroot y - 1 = 1/squareroot y - 1 it is continuous for, y > 1 where y_0 = 2 > 1 so there exist unique solution for initial value problem dy/dx = squareroot x - y f is continuous on region, y greterthanorequalto x y(1) = 1 1 greterthanorequalto 1, so (1,1) with in region R so there exist solution to this initial value problem f_y = (- 1)/2 squareroot x - y f_y is continuous on region, y > x (1,1) is not in region y > x so this initial value problem does not have unique solution